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I want to find this functional derivative:

$$\dfrac{\delta \int d^d x'[\nabla_{x'} \phi(\vec{x}')]^2}{\delta \phi(\vec{x})} = \int d^d x' \left(\dfrac{\delta \nabla_{x'} \phi(\vec{x}')}{\delta \phi(\vec{x})}\cdot \nabla_{x'} \phi(\vec{x}') +\nabla_{x'} \phi(\vec{x}')\cdot \dfrac{\delta \nabla_{x'} \phi(\vec{x}')}{\delta \phi(\vec{x})}\right)$$

According to the book I have, the first term of the integral is:

$$\int d^d x' \dfrac{\delta \nabla_{x'} \phi(\vec{x}')}{\delta \phi(\vec{x})}\cdot \nabla_{x'} \phi(\vec{x}') = -\int d^d x' \dfrac{\delta\phi(\vec{x}')}{\delta\phi(\vec{x})} \nabla^2_{x'} \phi(\vec{x}') \tag{1}$$

With that result and with $\dfrac{\delta\phi(\vec{x}')}{\delta\phi(\vec{x})}=\delta(\vec x -\vec x ')$, I get $-\nabla^2 \phi(\vec{x})$, but I cannot see where the minus comes from in $(1)$. In fact, why does the Laplacian appear in $(1)$?

David
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    Just integration by parts...after having interchanged the symbol of functional derivative with the one of gradient. – V. Moretti Nov 18 '14 at 07:00
  • @ValterMoretti Ok, I will do that – David Nov 18 '14 at 07:00
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    I voted to close this as purely a math question, but I think the current practice is to leave open these sorts of questions if there is some physics context. I would be willing to retract my close vote if you add a few sentences to explain the physics context of this particular integral. –  Nov 18 '14 at 14:23

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